We have 2 aircraft
plane A weighs 300 Tonnes and plane B weighs 200 Tonnes.
Both are identical and have the same wing with the same Lift drag ratio of 17 similar to a 747.
Both aircraft are cruising straight and level at 400 Km/h.
Plane A at 300 tonnes lift has a drag of 300 / 17 which equals 17.65 units of drag.
Plane B at 200 tonnes lift has a drag of 200 / 17 which equals 11.76 units of drag.
So plane A has 1.5 times more drag than Plane B. at this speed.
Plane A has a momentum of Mass x Velocity which is 300 x 400 = 120,000.
Plane B has a momentum of 200 x 400 = 80,000
So again we have plane A with 1.5 times the momentum of plane B.
So all is nicely in equilibrium with both planes nicely in straight and level flight.
Now we want to reduce the speed of both aircraft to 300 kmh.
Plane A has a kinetic energy of ½ x 300 x (400 x 400) = 24,000,000
At 300 kmh it will have KE of ½ x 300 x (300 x 300) = 13,500,000.
This means plane A has to lose 10,500,000 units of KE.
Plane B has a KE of ½ x 200 x (400 x 400) = 16,000,000
At 300 Kmh it will be ½ x 200 x (300 x 300) = 9,000,000
So plane B has to lose 7,000,000 units of KE.
Plane A has to lose 1.5 times the energy as plane B.
So as long as the lift drag ratio stays the same between the 2 velocities and the amount of lift needed, both aircraft will slow down at exactly the same rate.
If the lift drag ratio alters due to plane configuration it will affect the rate of deceleration between the 2 aircraft.
BUT maximum lift drag ratio is independent of the weight of the aircraft, the area of the wing, or the wing loading.
i‘ve tried to keep this simple so forgive the rubbish math’s, but I thought I’d give you all something to pull apart for a while.
GB